There is this isomorphism in my notes but there is no explanation. So I tried to reason myself but still not convincing enough, or my reasoning may even be wrong. I will appreciate if anyone is willing to lend some helps.


I have heard that we can achieve that by putting $i$ into $X$ in the denominator, but if we do that than the denominator will be $0$, so $\mathbb{R}[X]/(0)=\mathbb{R}[X]$? Then how can it be isomorphic to $\mathbb{C}$?

Thanks so much!


When you quotient a polynomial ring by (the ideal generated by) a polynomial, you automatically get a root to it. In this case, the element represented by $X$ satisfies $X^2 + 1$ (just by definition).

Using the polynomial division algorithm, you can write any element $f$ of $\mathbb{R}[X]$ as $f = q(X^2 + 1) + r$, where $r$ has degree $1$ or $0$. In the quotient, $f$ and $r$ represent the same element.

So an arbitrary element of your quotient has the form $aX + b$. If you write out what happens when you multiply and then reduce two such expressions, you recover the multiplication equation you get for $\mathbb{C}$.


Hint: the evaluation map $\,\Bbb R[x] \to \Bbb C\,$ where $\,f(x)\mapsto f(i)\,$ is onto, and its kernel is generated by the minimal polynomial of $\,i,\,$ i.e. $\,x^2+1,\,$ since $\,\Bbb R[x]\,$ is Euclidean so a PID. Therefore $\ \Bbb R[x]/(x^2+1)\cong \Bbb C\ $ by the first isomorphism theorem.


$\newcommand{\C}{\mathbb{C}}\newcommand{\R}{\mathbb{R}} \mathbb{C}$ is often described as being constructed from $\mathbb{R}$ by freely adjoining an element $i$ satisfying $i^2 = -1$ (equivalently, $i^2 + 1 = 0$). This isomorphism is one way of saying that formally, since the polynomial ring freely adjoins a new element, and then quotienting forces the equation $x^2 + 1 = 0$.

What does this mean, precisely? The following two facts are the core of it:

Proposition. Let $R$, $S$ be any rings. Then ring homomorphisms $R[x] \to S$ correspond to pairs of a homomorphism $R \to S$ (the image of the constant polynomials), together with an element of $S$ (the image of $x$).

This formalises the idea that $R[x]$ freely adjoins an element to $R$ — any time you have a ring with a map of $R$ and a chosen element, these extend uniquely to a map from $R[x]$, since the ring homomorphism conditions determine what the image of any polynomial must be once you know the images of the variable and the coefficients.

Proposition. Let $R$ be a ring, $a \in R$ any element. Then for any other ring $S$, homomorphisms $R/(a) \to S$ correspond to homomorphisms $R \to S$ that send $a$ to $0$.

This similarly formalises the idea that quotienting by $(a)$ is exactly forcing the equation $a = 0$.

Putting these together, $\R[x]/(x^2 + 1)$ is “freely adding an element with $x^2 = -1$ to $\R$”; so this explains intuitively why you should expect it to be isomorphic to $\C$.

To construct the isomorphism, put together the facts above. For any ring $S$, a homomorphism $\R[x]/(x^2+1) \to S$ is determined by specifying a homomorphism from $\R$, together with an element $a$, such that $a^2 + 1 = 0$; and this extension is unique, so to check that two homomorphisms out of $\R[x]/(x^2+1)$ agree, it’s enough to check what they do on $\R$ and $x$. This fact lets you easily construct a map $\R[x]/(x^2+1) \to \C$; then you can explicitly give a map $\C \to \R[x]/(x^2+1)$, and check it’s a homomorphism; and then you can use this fact again in checking that the maps are mutually inverse.


Everyone else has given very good answers, so let me just make a few hand-wavey remarks:

  • Would you be happy if I wrote $\mathbb{R}[i] = \mathbb{C}$? (I don't mind if you change $=$ to $\cong$ - this statement is a little imprecise anyway.) What properties does $i$ satisfy?
  • I think about the ring $\mathbb{R}[X]/(X^2 + 1)$, informally, as the ring generated from $\mathbb{R}$ by adding in an extra element $X$, subject to the relation that $X^2 + 1 = 0$, and nothing else. Can you make this statement precise? (What does a general element of $\mathbb{R}[X]/(X^2 + 1)$ look like? How do two such elements multiply?)
  • Bonus question 1: what is the dimension of $\mathbb{R}[i]$ as an $\mathbb{R}$-vector space? What about $\mathbb{R}[X]/(X^2 + 1)$? Can you write down bases?
  • Can you profitably interpret the two rings $\mathbb{R}[X]/(X^2 + 1)$ and $\mathbb{R}[i]$ as being the same thing, by interpreting $X$ as $i$? Can you write down an isomorphism making this statement precise? (Can you show it's (a) injective, (b) surjective, (c) a homomorphism of rings?)
  • Bonus question 2: can you see why the two rings are still isomorphic when you interpret $X$ as $-i$? Can you see why interpreting $X$ as $1+i$ doesn't yield an isomorphism?

I was surprised that nobody added an argument using the first isomorphism theorem and elementary means.

Define $$f: \mathbb{R}[X] \to \mathbb{C}: P \mapsto P(i)$$

It is easy to see that this is a ring epimorphism.

It is also easy to see that $(X^2 +1)\subseteq ker f$. For the other inclusion, assume $P\in ker f$ and consider $P$ as a complex polynomial. Then $P(i) =0$ means that $(X-i)|P$. Note that $P(-i) =0$, so $(X+i)|P$ as well.

It follows that $(X^2+1)|P$ in $\mathbb{C}[X]$. A moment of consideration tells us that then $(X^2+1)|P$ in $\mathbb{R}[X]$ as well which means $P \in (X^2 +1)$.


Consider $\mathbb{R}[X]/(x^2+1)\ni a+bX\mapsto a+i b\in \mathbb{C}$.


The polynomial $X^2+1$ is irreducible of degree 2.

Thus, the ideal $(X^2+1)$ is maximal in the ring $\Bbb R[X]$.

Thus, the quotient $\Bbb R[X]/(X^2+1)$ is a field extension of degree $2$ over $\Bbb R$.

How many are there?

  • $\begingroup$ How many...? :) $\endgroup$
    – wxu
    Jun 7 '13 at 16:30
  • $\begingroup$ @wxu : don't you know ?! :) $\endgroup$ Jun 7 '13 at 16:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.